Really annoyed me that the kiddos voted to not switch, and they got the car. Getting children to understand that 1/3 is smaller than 1/2 is hard enough. I’m going to program my own evil version that only wins on switch the first time next time I do this.
Change the problem from 3 doors to a million. Kids pick a door, and the host opens 999,998 doors, leaving theirs and one other door closed. One of the closed doors is the winner. Do they want to switch now?
This is a good way to think about it, because the three door version makes it almost seem like the host is guessing. But if you pick door #8 and the host eliminates every door except #8 and, for some reason, #682,025, are you more confident in 8 or 682,025?
That’s a good idea. I went with redoing the sim a few times and then also doing stuff with a plinko game, which seemed to help them get it.
The first couple plinko runs also were really annoying for bellcurve purposes tho.
I explained this to my programmer colleagues. My explanation was lacking so they wrote a numerical simulation and ran it a few million times to be sure.
They were convinced.
LOL I did this too when I first heard the solution.
I was like “no, that cant be right”. Then ran the test and was surprised by the result.
It’s called Monte Carlo method, I presume.
I talked a little about Monte Carlo methods too - told them about a FORTRAN (!) program I had to write in college. Finding pi by throwing darts at a dart board and looking at the ratio of hits to misses to determine area…
Have you played the Cliff Hangers music to those children, because if so I love you.
Oh this makes it so much more intuitive. I’m gonna use that in the future
My favorite trivia about the Monty Hall Problem is that Monty Hall doesn’t understand the Monty Hall Problem.
The issue is that this isn’t a fraction problem, it’s a statistics problem. You need sample size. You need to teach it with M&Ms or something small, and let the kids choose to switch or not switch and play 20 rounds each with that choice, then compare piles.
I don’t care to look it up, cause that ain’t the point of social media. What is the Monty Hall problem?
3 doors.
1 is a car. 2 are goats.
Contestant picks a door. The host then shows one of the remaining doors containing a goat, then asks if you want to change your door.
What do you do?
That’s a fun sim!
Lol! I played 12 times in a row, and always stayed. I got the car 6 times (not in a row). I know that it’s supposed to be a 1/3 vs 2/3 probability, but it sure does seem like 50/50 in the long run.
It’s not, it approaches 1/3.
12 isn’t quite enough goes to get a good approximation.
Oh, I know…but I played another 5 rounds afterwards and now I’m ahead by 2 on cars. I think I should buy a lottery ticket on my way home tonight. I’m shittin’ clovers right now. Lol!
I managed to get 3 wins and 17 losses by staying lol. I’ll hold off on those lottery tickets.
It was hard to picture looking at it in one direction for me as a kid, but after thinking about it for about 15 minutes I looked at it from the “opposite view” and for some reason it clicked. (Not from the players view, but rather the hosts view)
They give you the chance to open a door 1/3. Then ask if you would like to change after opening another door to show a goat. Your first thought is that, well it’s a 50-50 chance the last door is the prize because there are only 2 left. so what would it matter, but really it is 2/3. As your chances of being right were 1/3 originally, and he knows where the car is, and has to eliminate one of the doors you don’t have. Making it so now 2/3 of the options are out if you choose a new door, or it stays at 1/3 if you stay the same. So you say “thank you for the extra 33% chance at winning” and always change to the second door
When the first door is revealed, obviously the probabilities are unchanged. The almighty protagonist of human exceptionalism must choose the path to alter the winds of the fates. Pray the great Apollo smiles upon thy being. For speaking the word “same” is to alter the universe of satyr sophistry.
Thanks I wasn’t sure if I understood it right.
I tried 25 times with staying
And 25 times with switching
Switching gave me a winning rate of 76% and was like “hmm… If I understand it the right way the chances are bigger if I switch right?”
Could you do it a few times, and then have them keep track of how often they win when switching vs not switching?
What helped me understand way back when I found out about it was using 3 cards, one car, two goats. Then have the student choosing not see the cards and you and the rest can see them standing on the other side. If the student never switches, he/she has 1/3 chance of picking a car. If the student always switches, it becomes obvious, especially if you see the cards, that to win he/she has to pick a goat, which is 2/3 chance. It makes it even more obvious with more goat cards. The students watching immediately see which card must be chosen for two situations - never swap or always swap.
